We first have to find in #1/4# the limited development of #Arcsin(x)#
So, we have to find #LD_1(1/4)# #sin^(-1)(x)#
Let #X=x-1/4#
We have to find #LD(0)# #sin^(-1)(X+1/4)#
also, #f'(X)=1/sqrt(1-(X+1/4)^2)=1/sqrt(1-X^2-1/2X-1/16)#
#=1/sqrt(15/16-(X^2+1/2X))=16/15*1/sqrt(1-15/16(X^2+1/2X))#
Let #Y=-15/16(X^2+1/2X)#
We have #f'(Y)=16/15*1/sqrt(1+Y)=16/15*(1+Y)^(-1/2)#
And we know a limited development of #(1+Y)^n#
We will only keep term of degree 0 and 1.
#LD_1(0)# #(1+Y)^n=1+nY+o(Y)#
Let replace #Y# by #-15/16(X^2+1/2X)# and #n=-1/2#
#LD_1(0)# #f'(X)=1+15/32(color(red)(cancel(color(black)(X^2)))+1/2X)+o(X)#
#=1+15/64X#
And #LD_(n+1)(0)# #f(X)=f(0)+intLD_n(0)f(X)dX + o(X^(n+1))#
#<=>LD_2(0)sin^(-1)(X)=color(red)(cancel(color(black)(sin^(-1)(0))))^(=0)+int(1+15/64X)dX+o(X^2)#
#LD_2(0)sin^(-1)(X)=X+15/128X^2+o(X^2)#
Then, finally, replace #X=x-1/4#, knowing that
we only want a #LD_1(1/4)f(x)#
#LD_2(1/4)sin^(-1)(x)=x-1/4+15/128(x-1/4)^2+o((x-1/4)^2)#
#LD_1(1/4)sin^(-1)(x)=x-1/4+color(red)(cancel(color(black)(15/128x^2)))-15/256x+15/4096+o(x)#
#=241/256x-1009/4096#
So, the local linearization of sin^(-1)(x) in #a=1/4# is
#y=241/256x-1009/4096#
